You are given two charges $q_0$ and $q$ situated on the z axis at $z=+a$ and $z=-a$. What are the monopole and dipole terms in the potential, when the distance from origin $r>>a$ if

a) $q=q_0$

b) $q=-q_0$

c) $q=0$

$d=2a$

$R_±=r^2+(d/2)^2∓d*r*\cos θ= r^2 [1∓d/r*\cosθ+d^2/(4r^2 )]$

$1/R_± =(1/r)*\frac{1}{√((1∓d/r*cos〖θ+d^2/(4r^2 ))}≈(1/r)*(1/\sqrt{1∓d/r*cosθ})≈$

$≈(1/r)*[1∓(-1/2)*d/r \cosθ]$ (1)

If $q=q0$:

$V(r)=k(q/R_+ +q/R_- )$

$1/R_+ +1/R_- =1/r*[1+⋯.]+1/r*[1-….]=2/r$

$V(r)=2kq/r$

So that this potential contains only the monopole term (1/r) and the dipole term is zero.

If $q=-q0$

$V(r)=k(q/R_+ -q/R_- )$

$1/R_+ -1/R_- =(1/r)*[1+1/2*d/r*\cos θ]-(1/r)*[1-1/2*d/r*\cos θ=$

$=(1/r)*(d/r)*\cos θ$

$V(r)=k q/r^2 (d*\cos θ)$

So that this potential contains only the dipole term $(1/r^2)$ and the monopole term is zero.

If $q=0$

$V(r)=k q/R_+ =k q/r[1+d/2r*\cos θ]$

Therefore this potential contains also the monopole term $(1/r)$ and the dipole term $(1/r^2)$.

If one looks at the equation (1) above it sees that this is just an approximate one. In fact there are two approximations made one after another. First it is neglected the term $1/r^2$ under the radical, the the radical is again written as a power series. Higher terms in $1/r^2$ need to be included in the last result for a better approximation. Therefore a quadrupole term does exist in all cases a, b and c.